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# The volume of a spherical balloon is increasing at the rate of 25 cm3/sec. Find the rate of change of its surface area at the instant when the radius is 5 cm

SHAIK AASIF AHAMED
6 years ago
Hello student,
Let r,V and S be radius,volume and surface area of the sphere at time t.
$v=(4/3) \pi r^{3} and S=4\pi r^{2}$
Rate of change of volume =(dv/dt)=25
$4\pi r^{2}dr/dt=25$
Rate of change of surface area=(ds/dt)=8$\pi r$(dr/dt)=8$\pi r$(25/4$\pi r^{2}$)=50/r
So rate of change of surface area when r is 5cm=(50/5)=10cm2/sec
Thanks and Regards
Shaik Aasif
Dev Desai
13 Points
2 years ago
Volume of sphere = (4/3)(pi)(r)^3
Surface area of sphere =(4)(pi)(r)^2

V/S = ((4/3)(pi)(r)^3)/((4)(pi)(r)^2) =>  r/3
Therefore, ((dv/dt)×3)/r= (ds/dt)
Hence, ds/dt = (25×3)/5 = 10 cm^3/sec